Answer :
The given expression is :
[tex]2f-\frac{1}{3}(2g-3f)-\frac{1}{3}g[/tex]Simplify by the BODMAS rule
B-Brackets, O- Of, D-Division, M-Multiplication, A-Add & S - SUbtract
Step 1 : Open the brackets
[tex]2f-\frac{1}{3}(2g-3f)-\frac{1}{3}g=2f-\frac{2}{3}g+\frac{3}{3}f-\frac{1}{3}g[/tex]Step 2 : Simplify the similar term together
[tex]\begin{gathered} 2f-\frac{1}{3}(2g-3f)-\frac{1}{3}g=2f+\frac{3}{3}f-\frac{1}{3}g-\frac{2}{3}g \\ 2f-\frac{1}{3}(2g-3f)-\frac{1}{3}g=2f+f-\frac{1}{3}g-\frac{2}{3}g \\ 2f-\frac{1}{3}(2g-3f)-\frac{1}{3}g=f(2+1)-g(\frac{1}{3}+\frac{2}{3}) \\ 2f-\frac{1}{3}(2g-3f)-\frac{1}{3}g=f(3)-g(\frac{3}{3}) \\ 2f-\frac{1}{3}(2g-3f)-\frac{1}{3}g=f(3)-g(1) \\ 2f-\frac{1}{3}(2g-3f)-\frac{1}{3}g=3f-g \end{gathered}[/tex]Answer : A) 3f - g